Obviously, \(\displaystyle \mathcal {K}_1^F g=\int _{\mathbb {R}^3}l_1(v,v_1)g(v_1)\mathrm {d}v_1\)  with
\(l_1(v,v_1)=\int _{\mathbb {S}^2}\frac{|(v-v_1)\cdot \omega |\mathcal {N}}{\sqrt{\mu _F(1-\mu _F)}} \frac{\sqrt{(\mu _F)_1(1-(\mu _F)_1)}}{\mu _1(1-\mu _1)}\mathrm {d}\omega ,\) 
where \((\mu _F)_1=\mu _F(v_1)\) . It follows immediately by (REF ) and (REF ) that
\(|l_1(v,v_1)|\,\leqslant \, C|v-v_1|\mu _F^{\alpha -1/2}(v)\mu _F(v_1)\,\leqslant \, \frac{C}{|v-v_1|}e^{-c|v-v_1|^2}.\) 
For \(\mathcal {K}_2^F\) , the Proposition 2.2 in [1]} indicates that
\(\begin{aligned}\mathcal {K}_2^F g(v)=&2\iint _{\mathbb {R}^3\times \mathbb {S}^2} \frac{|(v-v_*)\cdot \omega |\mathcal {N}}{\sqrt{\mu _F(1-\mu _F)}}\frac{\sqrt{\mu _F^\prime (1-\mu _F^\prime )}}{\mu ^\prime (1-\mu ^\prime )}g^\prime \mathrm {d}\omega \mathrm {d}v_*\\=&2\iint _{\mathbb {R}^3\times \mathbb {S}^2} |(v-v_*)\cdot \omega |\mu _*^\prime \frac{\sqrt{\mu _F^\prime }}{\sqrt{\mu _F}}\frac{\sqrt{1-\mu _F^\prime }(1-\mu )(1-\mu _*)}{\sqrt{1-\mu _F}(1-\mu ^{\prime })}g^{\prime } \mathrm {d}\omega \mathrm {d}v_*,\end{aligned}\) 
Noting that \(1-\mu \)  and \(1-\mu _F\)  are both bounded and
\(\mu (v)\sim e^{-\frac{|v-\mathfrak {u}|^2}{2T}},\quad \mu _F(v) \sim e^{-\frac{|v|^2}{2T_m}},\) 
then by a similar argument as Lemma 2.3 in [2]}, we can write \(\mathcal {K}_2^F g\)  as
\(\mathcal {K}_2^F g(v)=\int _{\mathbb {R}^3}l_2(v,v_1)g(v_1)\mathrm {d}v_1,\) 
with
\(\begin{aligned}|l_2(v,v_1)|\leqslant \frac{C}{|v_1-v|}e^{-c|v_1-v|^2}.\end{aligned}\) 
The estimates (REF ) and (REF ) implies (REF ).
